/**
 * @param {number[]} nums
 * @return {number}
 */
const findDuplicate = function (nums) {
  const n = nums.length
  let ans = 0
  let bit_max = 31
  while ((n - 1) >> bit_max == 0) {
    bit_max -= 1
  }
  for (let bit = 0; bit <= bit_max; ++bit) {
    let x = 0,
      y = 0
    for (let i = 0; i < n; ++i) {
      if ((nums[i] & (1 << bit)) != 0) {
        x += 1
      }
      if (i >= 1 && (i & (1 << bit)) != 0) {
        y += 1
      }
    }
    if (x > y) {
      ans |= 1 << bit
    }
  }
  return ans
}

// another

/**
 * @param {number[]} nums
 * @return {number}
 */
const findDuplicate = function (nums) {
  const len = nums.length
  if (len > 0) {
    let slow = nums[0]
    let fast = nums[nums[0]]
    while (slow !== fast) {
      slow = nums[slow]
      fast = nums[nums[fast]]
    }
    slow = 0
    while (slow !== fast) {
      slow = nums[slow]
      fast = nums[fast]
    }
    return slow
  }
  return -1
}

// another

/**
 * @param {number[]} nums
 * @return {number}
 */
const findDuplicate = function (nums) {
  let n = nums.length - 1,
    res = 0
  for (let p = 0; p < 32; ++p) {
    let bit = 1 << p,
      a = 0,
      b = 0
    for (let i = 0; i <= n; ++i) {
      if (i > 0 && (i & bit) > 0) ++a
      if ((nums[i] & bit) > 0) ++b
    }
    if (b > a) res += bit
  }
  return res
}

// another

/**
 * @param {number[]} nums
 * @return {number}
 */
const findDuplicate = function (nums) {
  const hash = {}
  for (let i = 0; i < nums.length; i++) {
    if (hash.hasOwnProperty(nums[i])) {
      return nums[i]
    } else {
      hash[nums[i]] = 1
    }
  }
}

/**

# File: FindDuplicate.py
# Author: Keith Schwarz (htiek@cs.stanford.edu)
#
# An algorithm for solving the following (classic) hard interview problem:
#
# "You are given an array of integers of length n, where each element ranges
#  from 0 to n - 2, inclusive.  Prove that at least one  duplicate element must
#  exist, and give an O(n)-time, O(1)-space algorithm for finding some
#  duplicated element.  You must not modify the array elements during this
#  process."
#
# This problem (reportedly) took CS legend Don Knuth twenty-four hours to solve
# and I have only met one person (Keith Amling) who could solve it in less time
# than this.
#
# The first part of this problem - proving that at least one duplicate element
# must exist - is a straightforward application of the pigeonhole principle.
# If the values range from 0 to n - 2, inclusive, then there are only n - 1
# different values.  If we have an array of n elements, one must necessarily be
# duplicated.
#
# The second part of this problem - finding the duplicated element subject to
# the given constraints - is much harder.  To solve this, we're going to need a
# series of nonobvious insights that transform the problem into an instance of
# something entirely different.
#
# The main trick we need to use to solve this problem is to notice that because
# we have an array of n elements ranging from 0 to n - 2, we can think of the
# array as defining a function f from the set {0, 1, ..., n - 1} onto itself.
# This function is defined by f(i) = A[i].  Given this setup, a duplicated
# value corresponds to a pair of indices i != j such that f(i) = f(j).  Our
# challenge, therefore, is to find this pair (i, j).  Once we have it, we can
# easily find the duplicated value by just picking f(i) = A[i].
#
# But how are we to find this repeated value?  It turns out that this is a
# well-studied problem in computer science called cycle detection.  The general
# form of the problem is as follows.  We are given a function f.  Define the
# sequence x_i as
#
#    x_0     = k       (for some k)
#    x_1     = f(x_0)
#    x_2     = f(f(x_0))
#    ...
#    x_{n+1} = f(x_n)
#
# Assuming that f maps from a domain into itself, this function will have one
# of three forms.  First, if the domain is infinite, then the sequence could be
# infinitely long and nonrepeating.  For example, the function f(n) = n + 1 on
# the integers has this property - no number is ever duplicated.  Second, the
# sequence could be a closed loop, which means that there is some i so that
# x_0 = x_i.  In this case, the sequence cycles through some fixed set of
# values indefinitely.  Finally, the sequence could be "rho-shaped."  In this
# case, the sequence looks something like this:
#
#     x_0 -> x_1 -> ... x_k -> x_{k+1} ... -> x_{k+j}
#                        ^                       |
#                        |                       |
#                        +-----------------------+
#
# That is, the sequence begins with a chain of elements that enters a cycle,
# then cycles around indefinitely.  We'll denote the first element of the cycle
# that is reached in the sequence the "entry" of the cycle.
#
# For our particular problem of finding a duplicated element in the array,
# consider the sequence formed by starting at position n - 1 and then
# repeatedly applying f.  That is, we start at the last position in the array,
# then go to the indicated index, repeating this process.  My claim is that
# this sequence is rho-shaped.  To see this, note that it must contains a cycle
# because the array is finite and after visiting n elements, we necessarily
# must visit some element twice.  This is true no matter where we start off in
# the array.  Moreover, note that since the array elements range from 0 to
# n - 2 inclusive, there is no array index that contains n - 1 as a value.
# Consequently, when we leave index n - 1 after applying the function f one
# time, we can never get back there.  This means that n - 1 can't be part of a
# cycle, but if we follow indices starting there we must eventually hit some
# other node twice.  The concatenation of the chain starting at n - 1 with the
# cycle it hits must be rho-shaped.
#
# Moreover, think about the node we encounter that starts at the entry of the
# cycle.  Since this node is at the entry of the cycle, there must be two
# inputs to the function f that both result in that index being generated.  For
# this to be possible, it must be that there are indices i != j with
# f(i) = f(j), meaning that A[i] = A[j].  Thus the index of the entry of the
# cycle must be one of the values that is duplicated in the array.
#
# There is a famous algorithm due to Robert Floyd that, given a rho-shaped
# sequence, finds the entry point of the cycle in linear time and using only
# constant space.  This algorithm is often referred to as the "tortoise and
# hare" algorithm, for reasons that will become clearer shortly.
#
# The idea behind the algorithm is to define two quantities.  First, let c be
# the length of the chain that enters the cycle, and let l be the length of the
# cycle.  Next, let l' be the smallest multiple of l that's larger than c.
# I claim that for any rho-shaped sequence l' defined above, that
#
#    x_{l'} = x_{2l'}
#
# The proof is actually straightforward and very illustrative - it's one of my
# favorite proofs in computer science.  The idea is that since l' is at least
# c, it must be contained in the cycle.  Moreover, since l' is a multiple of
# the length of the loop, we can write it as ml for some constant m.  If we
# start at position x_{l'}, which is inside the loop, then take l' more steps
# forward to get to x_{2l'}, then we will just walk around the loop m times,
# ending up right back where we started.
#
# One key trick of Floyd's algorithm is that even if we don't explicitly know l
# or c, we can still find the value l' in O(l') time.  The idea is as follows.
# We begin by keeping track of two values "slow" and "fast," both starting at
# x_0.  We then iteratively compute
#
#    slow = f(slow)
#    fast = f(f(fast))
#
# We repeat this process until we find that slow and fast are equal to one
# another.  When this happens, we know that slow = x_j for some j, and
# fast = x_{2j} for that same j.  Since x_j = x_{2j}, we know that j must be at
# least c, since it has to be contained in the cycle.  Moreover, we know that j
# must be a multiple of l, since the fact that x_j = x_{2j} means that taking j
# steps while in the cycle ends up producing the same result.  Finally, j must
# be the smallest multiple of l greater than c, since if there were a smaller
# multiple of l greater than c then we would have reached that multiple before
# we reached j.  Consequently, we must have that j = l', meaning that we can
# find l' without knowing anything about the length or shape of the cycle!
#
# To complete the construction, we need to show how to use our information
# about l' to find the entry to the cycle (which is at position x_c).  To do
# this, we start off one final variable, which we call "finder," at x_0.  We
# then iteratively repeat the following:
#
#   finder = f(finder)
#   slow   = f(slow)
#
# until finder = slow.  We claim that (1) the two will eventually hit each
# other, and (2) they will hit each other at the entry to the cycle.  To see
# this, we remark that since slow is at position x_{l'}, if we take c steps
# forward, then we have that slow will be at position x_{l' + c}.  Since l' is
# a multiple of the loop length, this is equivalent to taking c steps forward,
# then walking around the loop some number of times back to where you started.
# In other words, x_{l' + c} = x_c.  Moreover, consider the position of the
# finder variable after c steps.  It starts at x_0, so after c steps it will be
# at position x_c.  This proves both (1) and (2), since we've shown that the
# two must eventually hit each other, and when they do they hit at position x_c
# at the entry to the cycle.
#
# The beauty of this algorithm is that it uses only O(1) external memory to
# keep track of two different pointers - the slow pointer, and then the fast
# pointer (for the first half) and the finder pointer (for the second half).
# But on top of that, it runs in O(n) time.  To see this, note that the time
# required for the slow pointer to hit the fast pointer is O(l').  Since l' is
# the smallest multiple of l greater than c, we have two cases to consider.
# First, if l > c, then this is l.  Otherwise, if l < c, then we have that
# there must be some multiple of l between c and 2c.  To see this, note that
# in the range c and 2c there are c different values, and since l < c at least
# one of them must be equal to 0 mod l.  Finally, the time required to find the
# start of the cycle from this point is O(c).  This gives a total runtime of at
# most O(c + max{l, 2c}).  All of these values are at most n, so this algorithm
# runs in time O(n).

def findArrayDuplicate(array):
    assert len(array) > 0

    # The "tortoise and hare" step.  We start at the end of the array and try
    # to find an intersection point in the cycle.
    slow = len(array) - 1
    fast = len(array) - 1

    # Keep advancing 'slow' by one step and 'fast' by two steps until they
    # meet inside the loop.
    while True:
        slow = array[slow]
        fast = array[array[fast]]

        if slow == fast:
            break

    # Start up another pointer from the end of the array and march it forward
    # until it hits the pointer inside the array.
    finder = len(array) - 1
    while True:
        slow   = array[slow]
        finder = array[finder]

        # If the two hit, the intersection index is the duplicate element.
        if slow == finder:
            return slow


*/
